Question

100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of ppm of $CaC{O}_3$ is: (molar mass of calcium bicarbonate is 162 g/mol and magnesium bicarbonate is 146 g/mol.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> JEE MAIN 2019

• A. $5000ppm$
• B. $10000ppm$
• C. $100ppm$
• D. $1000ppm$

Answer 1

The correct option is B $10000ppm$

$Ca(HC{O}_3{)}_2\to CaC{O}_3+H_{2}O+C{O}_2$
Now, this total amount of calcium carbonate formed is to be measured by taking into consideration both calcium as well as magnesium bicarbonate.
Thus, according to the data given we have to find the total degree of hardness which is given by,
$n_{eq}.CaC{O}_3=n_{eq}.Ca(HC{O}_3{)}_2+n_{eq}.Mg(HC{O}_3{)}_2$
$\frac{w}{100}\times 2=\frac{0.81}{162}\times 2+\frac{0.73}{146}\times 2$
$w=1g$
Thus, 1 g of calcium carbonate is present in 100 mL and in terms of part per million in 100 mL it is:
$\Rightarrow \frac{1}{100}\times {10}^6$
$\Rightarrow {10}^{4}ppm=10000ppm$
Thus, the correct answer is option B) 10000 ppm.